Re: DEB_BUILD_OPTIONS=parallel=n work not as expected

[Erik Schanze <schanzi_@gmx.de>]

Hi,

[Sorry for crossposting to debian-dpkg, but perhaps they could clarify this.
Orig. post was: http://lists.debian.org/debian-mentors/2009/12/msg00075.html]

Erik Schanze <schanzi_@gmx.de>:
> I added to my package "gif2png" the support of
> DEB_BUILD_OPTIONS=parallel=n, as you can see in rules file.
> 
> The output during build is:
> ...
> # Add here commands to compile the package.                                                              
> /usr/bin/make  --jobserver-fds=3,4 -j
> ...
> 
> I don't know who changed $(MAKEFLAGS) from "-j3" to "--jobserver-fds=3,4 -j" and
> why the number "3" was not used after "-j".
> 
> Could anybody please give an explanation?
> Is there something broken or did I miss something?
> 
As far as I understand, the substitution comes from "make" itself
and it occurs in every sub-make during parallel build. 
So it seems the "make" call in build target of rules file is a
sub-make call, right?

But is it the intended behaviour, that the make call in "build" target will
run with option "-j" without any number?
Because this will end in a build run with unlimited jobs, or did I miss smth.?

I expected that "make" call in "build" target will be executed with
"-j <n>" with <n> from "DEB_BUILD_OPTIONS=parallel=<n>".
If I use any other name than "MAKEFLAGS" for the variable, it works
as expected. There is the key?


A bit puzzled,

Erik


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